Book edition 9th

Author(s) R. Kent Nagle, Edward B. Saff, Arthur David Snider

Pages 616 pages

ISBN 9780321977069

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Short Answer

Consider the differential equation $\frac{d y}{d x} = x + s i n y$
⦁ A solution curve passes through the point $(1, \frac{π}{2})$ . What is its slope at this point?
⦁ Argue that every solution curve is increasing for $x > 1$ .
⦁ Show that the second derivative of every solution satisfies $\frac{d^{2} y}{{dx}^{2}} = 1 + x \cos y + \frac{1}{2} \sin 2 y .$
⦁ A solution curve passes through (0,0). Prove that this curve has a relative minimum at (0,0).

⦁ The slope at the point 2.

⦁ Yes, every solution curve is increasing for x > 1.

⦁ The second derivative of every solution satisfies the given equation.

⦁ Yes, the curve has a minimum at (0,0)

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

1(a): Find the slope of solution curve at 1,π2

Slope is given by $\frac{dy}{dx}$

So, slope of the solution curve at $(1, \frac{π}{2})$ is

$\frac{dy}{dx} |(1, \frac{π}{2}) = 1 + \sin (1, \frac{π}{2}) = 1 + 1 = 2$

Hence, the slope at the point is 2.

2(b): Compute dydx for x > 1

Since, $\frac{dy}{dx} = x + \sin y$ for all $y \in ℝ$

Then, $x + \sin y > 1$ , for all $x > 1$ .

i.e., $\frac{dy}{dx} > 1 > 0$ , for all $x > 1$ , $y \in ℝ$ .

Hence, from first derivative test, every solution curve is increasing for $x > 1$ .

3(c): Determine the second derivative.

Here

$\frac{dy}{dx} = x + \sin y$

Differentiate both sides with respect to x

$\frac{d^{2} y}{{dx}^{2}} = 1 + \cos y (\frac{dy}{dx}) = 1 + \cos y (x + \sin y) = 1 + x \cos y + \sin y \cos y = 1 + x \cos y + \frac{1}{2} \sin 2 y$

So, the second derivative of every solution satisfies the given equation.

4(d): Find second derivative at (0,0)

Since, $x + \sin y = 0$ at $(0, 0)$

we get, $\frac{dy}{dx} = 0$ at $(0, 0)$

thus, $(0, 0)$ is a critical point.

Also, from part (c) , $\frac{d^{2} y}{{dx}^{2}} = 1 + x \cos y + \frac{1}{2} \sin 2 y$

$\frac{d^{2} y}{{dx}^{2}} = 1 > 0$ at $(0, 0)$

From second derivative test, $(0, 0)$ is a point of relative minimum.

Therefore, the curve has a minimum at $(0, 0)$

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Fundamentals Of Differential Equations And Boundary Value Problems

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Short Answer

Step by Step Solution

1(a): Find the slope of solution curve at 1,π2

2(b): Compute dydx for x > 1

3(c): Determine the second derivative.

4(d): Find second derivative at (0,0)

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Fundamentals Of Differential Equations And Boundary Value Problems

Answers without the blur.

Short Answer

Step by Step Solution

1(a): Find the slope of solution curve at 1,π2

2(b): Compute dydx for x > 1

3(c): Determine the second derivative.

4(d): Find second derivative at (0,0)

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